# Complete Quadrilateral

## Contents

## Complete quadrilateral

Let four lines made four triangles of a complete quadrilateral. In the diagram these are One can see some of the properties of this configuration and their proof using the following links.

## Radical axis

Let four lines made four triangles of a complete quadrilateral. In the diagram these are

Let points and be the orthocenters of and respectively.

Let circles and be the circles with diameters and respectively. Prove that Steiner line is the radical axis of and

**Proof**

Let points and be the foots of perpendiculars and respectively.

Denote power of point with respect the circle

Therefore power of point with respect these three circles is the same. These points lies on the common radical axis of and Steiner line is the radical axis as desired.

**vladimir.shelomovskii@gmail.com, vvsss**

## Areas in complete quadrilateral

Let complete quadrilateral be given . Let and be the midpoints of and respectively.

Prove that where is the area of

**Proof**

Let and be the projections of and respectively onto

Let and be the circles with diameters and respectively. Similarly

Let common radical axes of and cross at point The power of the point with respect and is the same, therefore

Let

Therefore

**vladimir.shelomovskii@gmail.com, vvsss**

## Newton–Gauss line

Let four lines made four triangles of a complete quadrilateral.

In the diagram these are

Let points and be the midpoints of and respectively.

Let points and be the orthocenters of and respectively.

Prove that Steiner line is perpendicular to Gauss line

**Proof**

Points and are the centers of circles with diameters and respectively.

Steiner line is the radical axis of these circles.

Therefore as desired.

**vladimir.shelomovskii@gmail.com, vvsss**

## Shatunov-Tokarev line

Let the complete quadrilateral ABCDEF be labeled as in the diagram. Quadrilateral is not cyclic.

Let points be the orthocenters and points be the circumcenters of and respectively.

Let bisector cross bisector at point Let bisector cross bisector at point

Prove that

a) points and lie on circumcircle of

b) line is symmetric to Steiner line with respect centroid of

I suppose that this line was found independently by two young mathematicians Leonid Shatunov and Alexander Tokarev in 2022. I would be grateful for information on whether this line was previously known.

**Proof**

a) Points and lies on bisector of points and lies on bisector of

circle

Similarly circle as desired.

b) Let and be midpoints of and respectively.

It is clear that is centroid of

(midline of trapezium (midline of trapezium is parallelogram.

Similarly one can prove that point the midpoint of is symmetric to with respect

Therefore line coincide with Steiner line and line is symmetric to Steiner line with respect and is parallel to this line.

**vladimir.shelomovskii@gmail.com, vvsss**